Q: Jill has 2.75 in nickels and quarters If he has 15 coins how many of each type does he have.?

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He has jillian.

Let X equal the number of quarters X * 25 is the value of the quarters ((X+8) * 5) is the value of the nickels25X + 5X + 40 = 610 so 30X + 40 = 610 .subtract 40 from both sides , divide both sides by 30X = 19There are 19 quarters and 27 nickels in the piggy bank

If you start with 30 nickels, you have $1.50. Every time you replace a nickel by a quarter, you gain 20 cents. $4.10 is $2.60 more than $1.50, so you need 13 quarters, leaving 17 nickels. Check: 13*0.25 + 17*0.05 = 3.25 + 0.85 = 4.10.

Each quarter has 5 nickels. 5 times 6.

1.508.025.060.0= 94.50

Chen has 8 nickels, 2 dimes, and 6 quarters.

The answer depends on the constraints of the problem.If you are required to have at least one of each of the named coins: 4 pennies, 2 nickels, 1 dime, and 3 quarters is 99 cents ($0.99).If you just have to pick from these, but not necessarily choose at least one of each: 4 pennies, 0 nickels, 9 dimes, 3 quarters is $1.69

The value of 5 quarters is 125 cents. Whether that is in dimes or whatever else, the value is always 125 cents. If, for some unknown and perverse reason, you want to use 5 quarters to buy nickels, then you'll get 5 nickels for each quarter, 25 nickels for all 5 quarters.

13 nickels 7 dimes

It's not possible to give a specific answer without knowing how the numbers of each coin are related. Otherwise there are hundreds of possible combinations; for example278 quarters, 0 dimes, 0 nickels277 quarters, 0 dimes, 5 nickels277 quarters, 1 dime, 3 nickels277 quarters, 2 dimes, 1 nickel276 quarters, 0 dimes, 10 nickelsand so on

The idea is to write two equations, one for the number of coins, one for the amount of money. Then solve the equations.Assuming "n" is the number of nickels, and "q" the number of quarters, the equations for the coins, of course, is quite simply: n + q = 64 And the equation for the money (I'll use cents; you can just as well use dollars if you prefer): 5n + 25q = 740 You can solve the first equation for "n", then replace that in the second equation.

Any of these four answers satisfies the question: Nickel--Dime--Quarter . 2 . . . . 7 . . . . 0 . 3 . . . . 5 . . . . 1 . 4 . . . . 3 . . . . 2 . 5 . . . . 1 . . . . 3